Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{x - 6}{x^3 - 3x^2 - 18x} \times \dfrac{-2x^3 - 24x^2 - 54x}{x - 8} $
Answer: First factor out any common factors. $r = \dfrac{x - 6}{x(x^2 - 3x - 18)} \times \dfrac{-2x(x^2 + 12x + 27)}{x - 8} $ Then factor the quadratic expressions. $r = \dfrac {x - 6} {x(x + 3)(x - 6)} \times \dfrac {-2x(x + 3)(x + 9)} {x - 8} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(x - 6) \times -2x(x + 3)(x + 9) } { x(x + 3)(x - 6) \times (x - 8)} $ $r = \dfrac {-2x(x + 3)(x + 9)(x - 6)} {x(x + 3)(x - 6)(x - 8)} $ Notice that $(x + 3)$ and $(x - 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-2x\cancel{(x + 3)}(x + 9)(x - 6)} {x\cancel{(x + 3)}(x - 6)(x - 8)} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $r = \dfrac {-2x\cancel{(x + 3)}(x + 9)\cancel{(x - 6)}} {x\cancel{(x + 3)}\cancel{(x - 6)}(x - 8)} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $r = \dfrac {-2x(x + 9)} {x(x - 8)} $ $ r = \dfrac{-2(x + 9)}{x - 8}; x \neq -3; x \neq 6 $